## 6th Edition 8.73

305113590
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### 6th Edition 8.73

Use the bond enthalpies in Tables 8.6 and 8.7 to estimate the reaction enthalpy for: CH4(g)+4Cl2(g)--->CCl4(g)+4HCl(g).

Can someone break down this step for me? I did 4(412)+4(0) - 4(338)- 4(431)=-1428 kJ/mol.
However, when I accounted for Cl2(g) being 242kJ/mol, I got the textbook answer of -460 kJ/mol. I thought Cl2 was 0.

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

### Re: 6th Edition 8.73

You're confusing standard enthalpy of formation and bond enthalpy. The standard enthalpy of formation of Cl2(g) is 0 because that is chlorine's standard state; however, we know that bond enthalpies are never 0 because breaking bonds require energy and forming bonds release energy. Therefore, you would need energy to break the Cl--Cl bond.

Aili Ye 4L
Posts: 58
Joined: Fri Sep 28, 2018 12:16 am

### Re: 6th Edition 8.73

As the other person mentioned, standard enthalpy of formation and bond enthalpy is different. Bond enthalpy cannot be zero since all bonds require energy to break.

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

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